Given f (2x + 1) = 8x + 74x2 + 4x + 2, find the range of F (x)

Given f (2x + 1) = 8x + 74x2 + 4x + 2, find the range of F (x)


∵ f (2x + 1) = 8x + 74x2 + 4x + 2, ∵ f (2x + 1) = 4 (2x + 1) + 3 (2x + 1) 2 + 1, that is, f (x) = 4x + 3x2 + 1, Let f '(x) = - 2 (x + 2) (2x-1) (x2 + 1) 2 = 0, the solution is x = - 2 or 12, when x ∈ (- ∞, - 2), f' (x) = - 2 (x + 2) (2x-1) (x2 + 1) 2 < 0, when x ∈ (- 2, 12), f '(x) = - 2 (x + 2) (2x -



Given the function f (x) = x & # 178; (AX + 1) defined on R, where a is a constant. If x = 1 is an extreme point of function y = FX, find the value of A. if y = FX is an increasing function on interval (0,2), find the value range of real number a


1. Derivative = 3ax ^ 2 + 2x is 0 at x = 1, so 3A + 2 = 0, a = - 2 / 3
2. Derivative = 3ax ^ 2 + 2x = x (3ax + 2), if a = 0, the problem is satisfied; if a is not 0, then the two are (0,, - 2 / 3a), and there is no intersection between (0.2) and this interval, so - 2 / 3A0
In general, a > = 0



The domain of two different functions f (x) = x2 + ax + 1 and G (x) = x2 + X + a (a is a constant) is R. if the range of F (x) and G (x) is the same, then a=___ I really can't understand the same range!


F (x) = x ^ 2 + ax + 1 = (x + A / 2) ^ 2 - A ^ 2 / 4 + 1 > = - A ^ 2 / 4 + 1g (x) = x ^ 2 + X + a = (x + 1 / 2) ^ 2 - 1 / 4 + 1 > = - 1 / 4 + A to make their ranges the same, then - A ^ 2 / 4 + 1 = - 1 / 4 + a = = > A = 1 or - 5 can be simply understood as the value range of Y